LeetCode: Binary Tree Inorder Traversal 解题报告

Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},
   1
    \
     2
    /
   3
return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

SOL:

包括递归与非递归方法：

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<Integer> inorderTraversal1(TreeNode root) {
12         List<Integer> ret = new ArrayList<Integer>();
13         rec(root, ret);
14         return ret;
15     }
16
17     public void rec(TreeNode root, List<Integer> ret) {
18         if (root == null) {
19             return;
20         }
21
22         rec(root.left, ret);
23         ret.add(root.val);
24         rec(root.right, ret);
25     }
26
27     public List<Integer> inorderTraversal(TreeNode root) {
28         List<Integer> ret = new ArrayList<Integer>();
29         if (root == null) {
30             return ret;
31         }
32
33         Stack<TreeNode> s = new Stack<TreeNode>();
34         TreeNode cur = root;
35
36         while (true) {
37             while (cur != null) {
38                 s.push(cur);
39                 cur = cur.left;
40             }
41
42             if (s.isEmpty()) {
43                 break;
44             }
45
46             cur = s.pop();
47             ret.add(cur.val);
48
49             cur = cur.right;
50         }
51
52         return ret;
53     }
54 }